# How do you solve 0.23x^2+6.5x+4.3<0 using a sign chart?

Nov 15, 2017

Solution : $- 27.58 < x < - 0.68 \mathmr{and} x | \left(- 27.58 , - 0.68\right)$

#### Explanation:

$0.23 {x}^{2} + 6.5 x + 4.3 < 0$

Comparing with standard quadratic equation $a {x}^{2} + b x + c = 0$

$a = 0.23 , b = 6.5 , c = 4.3$ Discriminant $D = {b}^{2} - 4 a c$ or

$D \approx 38.29$ If discriminant positive, we get two real solutions,

Quadratic formula: $x = \frac{- b \pm \sqrt{D}}{2 a}$or

$x = \frac{- 6.5 \pm \sqrt{38.29}}{2 \cdot 0.23} \therefore x \approx - 27.58 , x \approx - 0.68$

$0.23 {x}^{2} + 6.5 x + 4.3 < 0$or

$f \left(x\right) = 0.23 \left(x + 27.58\right) \left(x + 0.68\right) < 0$ .

Critical points are $x \approx - 27.58 , x \approx - 0.68$

Sign chart: When $x < - 27.58$ sign of $f \left(x\right)$ is  (-) * (-) = (+) ; > 0

When $- 27.58 < x < - 0.68$ sign of $f \left(x\right)$ is  (+) * (-) = (-) ; < 0

When $x > - 0.68$ sign of $f \left(x\right)$ is  (+) * (+) = (+) ; > 0

Solution : $- 27.58 < x < - 0.68 \mathmr{and} x | \left(- 27.58 , - 0.68\right)$

graph{0.23x^2+6.5x+4.3 [-160, 160, -80, 80]}

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